The solution of the differential equation (y+xxy(x+y))dx+(yxy(x+y)−x)dy=0 is

ydx−xdy+xxy(x+y)dx+yxy(x+y)dy=0ydx−xdy+(x+y)xy(xdx+ydx)=0ydx−xdyy2+xy+1xydx2+y22=0dx2+y22+dxyxy+1xy=0⇒x2+y22+2tan−1⁡xy=c

The solution of the differential equation (y+xxy(x+y))dx+(yxy(x+y)−x)dy=0 is

ydx−xdy+xxy(x+y)dx+yxy(x+y)dy=0ydx−xdy+(x+y)xy(xdx+ydx)=0ydx−xdyy2+xy+1xydx2+y22=0dx2+y22+dxyxy+1xy=0⇒x2+y22+2tan−1⁡xy=c

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$The solution of the differential equation (y + x \sqrt{x y} (x + y)) d x + (y \sqrt{x y} (x + y) - x) d y = 0 is$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{x^{2} + y^{2}}{2} + 2 \tan^{- 1} \sqrt{\frac{x}{y}} = c$

$x^{2} + y^{2} + \tan^{- 1} \frac{x^{2}}{y} = c$

$\frac{x^{2} y^{2}}{\sqrt{2}} + \tan^{- 1} \frac{x}{y} = c$

$\frac{x^{2} + y^{2}}{2} + \tan^{- 1} \frac{y}{x^{2}} = c$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} y d x - x d y + x \sqrt{x y} (x + y) d x + y \sqrt{x y} (x + y) d y = 0 \\ y d x - x d y + (x + y) \sqrt{x y} (x d x + y d x) = 0 \\ \frac{y d x - x d y}{y^{2}} + (\frac{x}{y} + 1) \sqrt{\frac{x}{y}} (d (\frac{x^{2} + y^{2}}{2})) = 0 \\ d (\frac{x^{2} + y^{2}}{2}) + \frac{d (\frac{x}{y})}{(\frac{x}{y} + 1) \sqrt{\frac{x}{y}}} = 0 \\ \Rightarrow \frac{x^{2} + y^{2}}{2} + 2 \tan^{- 1} \sqrt{\frac{x}{y}} = c \end{array}$